The problem asks to prove the inequality:

[ [x] + [y] \leq [x + y] \leq [x] + [y] + 1, ]

where $x$ and $y$ are real numbers, and $[x]$ denotes the greatest integer less than or equal to $x$ (also called the floor function of $x$).

Proof:

Step 1: Proving $[x] + [y] \leq [x + y]$

• Let $x = a + r$ and $y = b + s$, where $a = [x]$, $b = [y]$, and $r, s \in [0, 1)$ (the fractional parts of $x$ and $y$).

  Then: $x + y = (a + b) + (r + s).$

• The greatest integer function satisfies: [ [x + y] = [a + b + (r + s)]. ]

  • If $r + s < 1$, then $[x + y] = a + b = [x] + [y]$.
  • If $r + s \geq 1$, then $[x + y] = a + b + 1 \geq [x] + [y]$.

Hence, in either case: [ [x] + [y] \leq [x + y]. ]

Step 2: Proving $[x + y] \leq [x] + [y] + 1$

• Using the same representation $x = a + r$ and $y = b + s$: $x + y = (a + b) + (r + s).$

• Clearly, $r + s < 2$, since $r, s \in [0, 1)$. Thus: $a + b \leq x + y < a + b + 2.$

  Taking the floor of $x + y$, we get: [ [x + y] \leq a + b + 1 = [x] + [y] + 1. ]

Conclusion:

Combining the two parts, we conclude: [ [x] + [y] \leq [x + y] \leq [x] + [y] + 1. ]

This completes the proof.

Would you like me to provide further clarifications or examples?

Related Questions:

1. What are some properties of the floor function that are useful for inequalities?
2. How does this inequality change if $x$ and $y$ are integers instead of real numbers?
3. What happens to the inequality if we use the ceiling function instead of the floor function?
4. Can you provide a numerical example to verify the inequality?
5. How does the fractional part of a number affect these kinds of inequalities?

Tip:

Remember, the floor function simplifies many proofs by reducing real numbers to integers while controlling the error with fractional parts.